Sine rule and ambiguous cases | Year 11

01 / LEARN

The idea, step by step

Use opposite pairs

Write side a opposite angle A, and similarly for b and c. The sine rule works for any triangle. It is useful when one opposite pair is known. A side next to A is not necessarily side a: a lies across the triangle.

a / sin A = b / sin B = c / sin C

Check the second angle

Inverse sine gives a principal angle; its supplement has the same sine. Test both using the angle sum and all given data. A sine greater than 1 means no triangle. A candidate is invalid if it leaves a nonpositive third angle. SSA may therefore give zero, one or two triangles.

A common trap

Keeping only the calculator inverse-sine answer.

Instead: For SSA, test the supplementary angle before deciding how many triangles exist.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

Given $A = 30^{\circ}$ , $B = 45^{\circ}$ , $a = 8$ cm, find b exactly.

$b = 8 sin 45^{\circ} / sin 30^{\circ}$ .
$b = 8 (\sqrt{2} / 2) / (1 / 2)$ .
Simplify without early rounding.

8 \sqrt{2} cm

Example 2

Worked example 2

Given $A = 30^{\circ}$ , $a = 8$ cm and $b = 8 \sqrt{3}$ cm, find possible B and C.

$sin B = b sin A / a = \sqrt{3} / 2$ .
B is 60° or 120°; C is 180°−30°−B.
Both possible third angles are positive.

(B, C) = (60^{\circ}, 90^{\circ})

(120^{\circ}, 30^{\circ})

Check your understanding

If $sin B = 0.5$ and B is inside a triangle, what angles must be tested?

Show answer

30° and 150°.

Supplementary angles have the same sine.

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

A=30°, B=90°, a=6 cm. Find b.
3 marks
A=45°, B=60°, a=10 cm. Find b exactly.
3 marks
A=30°, a=5 cm, b=12 cm. Decide how many triangles exist.
3 marks
Given $A = 30^{\circ}$ , $a = 5$ cm, $b = 5 \sqrt{2}$ cm, find all possible B,C pairs.
3 marks
A=120°, a=10 cm, b=5 cm. Find B and C to one decimal place.
3 marks
A survey triangle has A=40°, B=65° and a=18 m. Find b to one decimal place and explain why b>a.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A=30°, B=90°, a=6 cm. Find b.
3 marks
$b = 6 sin 90^{\circ} / sin 30^{\circ} = 12$ .
12 cm
A=45°, B=60°, a=10 cm. Find b exactly.
3 marks
$b = 10 (\sqrt{3} / 2) / (\sqrt{2} / 2) = 5 \sqrt{6}$ .
$5 \sqrt{6} cm$
A=30°, a=5 cm, b=12 cm. Decide how many triangles exist.
3 marks
$sin B = 12 (1 / 2) / 5 = 1.2 > 1$ .
None.
Given $A = 30^{\circ}$ , $a = 5$ cm, $b = 5 \sqrt{2}$ cm, find all possible B,C pairs.
3 marks
$sin B = \sqrt{2} / 2$ .
Test 45° and 135°, then subtract each from 150°.
$(45^{\circ}, 105^{\circ})$ or $(135^{\circ}, 15^{\circ})$
A=120°, a=10 cm, b=5 cm. Find B and C to one decimal place.
3 marks
$sin B = \sqrt{3} / 4$ .
B≈25.6589°; its supplement is incompatible with A=120°.
C≈34.3411°.
B=25.7°, C=34.3°.
A survey triangle has A=40°, B=65° and a=18 m. Find b to one decimal place and explain why b>a.
3 marks
$b = 18 sin 65^{\circ} / sin 40^{\circ} \approx 25.3793$ .
65° exceeds 40°, so its opposite side is longer.
b≈25.4 m; b faces the larger angle.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: triangle rules, circular measure and trigonometric functions; school sequencing may vary.

Official curriculum source ↗

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